What is the value of sin(-pi/6)+cos(-pi/3) ?

The value of sin(-pi/6)+cos(-pi/3) is 0

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sin(-pi/6)+cos(-pi/3)

sin (- \frac{π}{6}) + cos (- \frac{π}{3})

0

Solution steps

sin (- \frac{π}{6}) + cos (- \frac{π}{3})

Use the following property:

cos (- x) = cos (x)

cos (- \frac{π}{3}) = cos (\frac{π}{3})

= sin (- \frac{π}{6}) + cos (\frac{π}{3})

Use the following property:

sin (- x) = - sin (x)

sin (- \frac{π}{6}) = - sin (\frac{π}{6})

= - sin (\frac{π}{6}) + cos (\frac{π}{3})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{1}{2}

= - \frac{1}{2} + \frac{1}{2}

Simplify

= 0

tan (\frac{5 π}{18})

tan^{2} (4 5^{\circ} + 1 6^{\circ})

\frac{( 7.4 \cdot 9.81 \cdot sin ( 6 5 ^{\circ} ) ) - 3 - ( 2 \cdot 9.81 )}{9.4}

- 3 sin (6 0^{\circ})

\frac{sin ( 3 0 ^{\circ} )}{sin ( 19. 5 ^{\circ} )}

What is the value of sin(-pi/6)+cos(-pi/3) ?
The value of sin(-pi/6)+cos(-pi/3) is 0